Question

A circuit consists of a series combination of 6.50 −kΩ and 5.00 −kΩ resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00 −kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.(a) What potentialdifference does the voltmeter measure across the 5.00-kΩ resistor? (b) What is the true potential difference across thisresistor when the meter is not present? (c) By what percentage isthe voltmeter reading in error from the true potentialdifference?

Answer 1

Part a)

`V = 16.95 Volts`

Part b)

`V = 21.74 V`

Part c)

`error = 22`%

Step-by-step explanation:

Part a)

When voltmeter of 10 k ohm is used to measure the voltage across the given resistor then we will have

`R_(eq) = (R_1 R_2)/(R_1 + R_2)`

`R_(eq) = (5 * 10)/(5 + 10)`

`R_(eq) = (50)/(15) = (10)/(3) k` ohm

now the current through the given cell is

`i = (V)/(r_1 + r_2)`

`i = (50)/(6.50 + (10)/(3))* 10^(-3)`

`i = 5.08 mA`

now voltage across 5 k ohm resistor is given as

`V = i R`

`V = (5.08 mA)((10)/(3)* 10^3)`

`V = 16.95 Volts`

Part b)

If there is not voltmeter connected across 5 k ohm

then current in the circuit is given as

`i = (50)/(6.5 + 5) * 10^(-3) A`

`i = 4.35 mA`

Now voltage across 5 k ohm resistor

`V = iR`

`V = 4.35 * 10^(-3) (5 * 10^3)`

`V = 21.74 V`

Part c)

percentage error in the reading is given as

`error = (V_1 - V_2)/(V_1) * 100`

`error = (21.74 - 16.95)/(21.74) * 100`

`error = 22`%